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I've been studying for my final exams, and I came across the following question:

If possible, give an example of a simple group $G$ with $n>1$ Sylow $p$-subgroups such that the order of $G$ does not divide $n!$. If not possible, briefly explain why.

Now, it didn't take me long to think of the following "numerical" example: a group of order $60=2^2\cdot3\cdot5$ where there are $3$ Sylow $2$-subgroups. My question is whether or not such a group exists, and if it does, is there any simple way to describe these groups based on knowing how the Sylow structure is built?

In trying to describe the above group, I think I showed that it can't exist. This is because there either have to be $4$ Sylow 3-subgroups or $10$ Sylow 3-subgroups. The second case leads to a contradiction since then there would have to be $k$ Sylow 5-subgroups such that $k(5-1)=30$, and likewise, in the first case, we find we'd have to have 13 Sylow 5-subgroups, also a contradiction. Is all of this correct and is there an easy way to determine whether such a group exists or not?

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Is ther ...there are $2$ Sylow $2$-subgroups....?? :-) – Babak S. Dec 15 '12 at 16:46
@BabakSorouh: I'm not sure I understand..?? Maybe I'm a bit slow today :) – Clayton Dec 15 '12 at 17:08
@Clayton: you are probably right - doesn't exist. As to your reasoning, if I did this right, $A_5$ has 10 Sylow-3 groups, so the argument you give for the contradiction (which I didn't quite get) can't be right. – gnometorule Dec 15 '12 at 17:17
It can't have $10$ Sylow $3$-subgroups if it has specifically $3$ Sylow 2-subgroups; that is what my argument is geared toward: the construction of a group of order $60$ with $3$ Sylow 2-subgroups. – Clayton Dec 15 '12 at 17:27
Sorry about the confusion; I was unaware. I knew it as a Sylow $p$-subgroup where $p$ was $2$, so naturally I substituted $2$ in for $p$. My apologies. – Clayton Dec 15 '12 at 18:06
up vote 2 down vote accepted

The answer is no.

Let $C$ be the set of sylow $p$ subgroups of $G$. It is easy to verify that the map $G\times C\rightarrow C$ that sends $(x,H)$ to $xHx^{-1}$ is an action of $G$ on $C$. By the orbit stabilizer theorem, we know that: $$\forall H\in C[|C||N(H)|=|G|]$$ Since $1<|C|$, we deduce that $|N(H)|<|G|$ for all $H\in C$....(1)

By sylow's theorem we know that the function $T_x:C\rightarrow C$ that sends $H$ to $xHx^{-1}$ is a permutation of $C$. Now consider the homomorphism $\phi:G\rightarrow S_C$ that sends $x$ to $T_x$.

Claim: $\phi$ is an injection.

Proof: We know that $\operatorname{Im}\phi\cong G/\ker\phi$. Thus, it suffices to show that $|\ker\phi|=1$. This is easy to see because $\ker\phi\unlhd G$, since $G$ is simple, therefore either $\ker\phi=\{e\}$ or $\ker\phi=G$. Let $H\in C$, it is easy to verify that $\ker\phi \leq N(H)$. Thus, if $\ker\phi=G$, we would conclude that $N(H)=G$. However, we know from (1) that $N(H)$ is a proper subgroup of $G$. Therefore, we finally deduce that $\ker\phi=\{e\}$.

Thus, $G$ is isomorphic to a subgroup of $S_C$ (The group of permutations of $C$). Hence, $|G|\,\Big|\,|S_C|!=n!$

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Thank you! Just to check, your line $|C|\cdot|N(H)|=|G|$ follows from the fact that the number of conjugate subgroups of $H$ is equal to the index of the normalizer? – Clayton Dec 17 '12 at 2:29
Yes. Which also follows from the orbit stabilizer theorem – Amr Dec 17 '12 at 10:13

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